On regular variation of entire Dirichlet series

Q3 Mathematics

Matematychni Studii Pub Date : 2023-01-16 DOI:10.30970/ms.58.2.174-181

P. Filevych, O. B. Hrybel

{"title":"On regular variation of entire Dirichlet series","authors":"P. Filevych, O. B. Hrybel","doi":"10.30970/ms.58.2.174-181","DOIUrl":null,"url":null,"abstract":"Consider an entire (absolutely convergent in $\\mathbb{C}$) Dirichlet series $F$ with the exponents $\\lambda_n$, i.e., of the form $F(s)=\\sum_{n=0}^\\infty a_ne^{s\\lambda_n}$, and, for all $\\sigma\\in\\mathbb{R}$, put $\\mu(\\sigma,F)=\\max\\{|a_n|e^{\\sigma\\lambda_n}:n\\ge0\\}$ and $M(\\sigma,F)=\\sup\\{|F(s)|:\\operatorname{Re}s=\\sigma\\}$. Previously, the first of the authors and M.M.~Sheremeta proved that if $\\omega(\\lambda)<C(\\rho)$, then the regular variation of the function $\\ln\\mu(\\sigma,F)$ with index $\\rho$ implies the regular variation of the function $\\ln M(\\sigma,F)$ with index $\\rho$, and constructed examples of entire Dirichlet series $F$, for which $\\ln\\mu(\\sigma,F)$ is a regularly varying function with index $\\rho$, and $\\ln M(\\sigma,F)$ is not a regularly varying function with index $\\rho$. For the exponents of the constructed series we have $\\lambda_n=\\ln\\ln n$ for all $n\\ge n_0$ in the case $\\rho=1$, and $\\lambda_n\\sim(\\ln n)^{(\\rho-1)/\\rho}$ as $n\\to\\infty$ in the case $\\rho>1$. In the present article we prove that the exponents of entire Dirichlet series with the same property can form an arbitrary sequence $\\lambda=(\\lambda_n)_{n=0}^\\infty$ not satisfying $\\omega(\\lambda)<C(\\rho)$. More precisely, if $\\omega(\\lambda)\\ge C(\\rho)$, then there exists a regularly varying function $\\Phi(\\sigma)$ with index $\\rho$ such that, for an arbitrary positive function $l(\\sigma)$ on $[a,+\\infty)$, there exists an entire Dirichlet series $F$ with the exponents $\\lambda_n$, for which $\\ln \\mu(\\sigma, F)\\sim\\Phi(\\sigma)$ as $\\sigma\\to+\\infty$ and $M(\\sigma,F)\\ge l(\\sigma)$ for all $\\sigma\\ge\\sigma_0$.","PeriodicalId":37555,"journal":{"name":"Matematychni Studii","volume":null,"pages":null},"PeriodicalIF":0.0000,"publicationDate":"2023-01-16","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"1","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"Matematychni Studii","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.30970/ms.58.2.174-181","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"Q3","JCRName":"Mathematics","Score":null,"Total":0}

引用次数: 1

Abstract

Consider an entire (absolutely convergent in $\mathbb{C}$) Dirichlet series $F$ with the exponents $\lambda_n$, i.e., of the form $F(s)=\sum_{n=0}^\infty a_ne^{s\lambda_n}$, and, for all $\sigma\in\mathbb{R}$, put $\mu(\sigma,F)=\max\{|a_n|e^{\sigma\lambda_n}:n\ge0\}$ and $M(\sigma,F)=\sup\{|F(s)|:\operatorname{Re}s=\sigma\}$. Previously, the first of the authors and M.M.~Sheremeta proved that if $\omega(\lambda)1$. In the present article we prove that the exponents of entire Dirichlet series with the same property can form an arbitrary sequence $\lambda=(\lambda_n)_{n=0}^\infty$ not satisfying $\omega(\lambda)

查看原文本刊更多论文

关于整个Dirichlet级数的正则变分

考虑一个完整的(在$\mathbb{C}$中绝对收敛的)Dirichlet级数$F$，其指数为$\lambda_n$，即形式为$F(s)=\sum_{n=0}^\infty a_ne^{s\lambda_n}$，并且对于所有$\sigma\in\mathbb{R}$，放入$\mu(\sigma,F)=\max\{|a_n|e^{\sigma\lambda_n}:n\ge0\}$和$M(\sigma,F)=\sup\{|F(s)|:\operatorname{Re}s=\sigma\}$。此前，第一个作者和M.M. Sheremeta证明了$\omega(\lambda)1$。本文证明了具有相同性质的整个狄利克雷级数的指数可以形成一个任意序列$\lambda=(\lambda_n)_{n=0}^\infty$不满足$\omega(\lambda)

本文章由计算机程序翻译，如有差异，请以英文原文为准。

求助全文

约1分钟内获得全文求助全文

来源期刊